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Have a look on the sketch

enter image description here

To get such a intensity distribution of light behind a slit we presuppose that all the light that moves on the two lines is coherent; of the same wavelength AND the same phase. Otherwise we shall get a blurred spot without fringes behind a slit.

To get the best interference pattern one has to use monochromatic light. Coherent light is not necessary and does not bring higher quality fringes. So I wondering, how the radiation gets coherent during the transition of a slit.

Edit after Anna's answer.

The usual light sources are not point-like sources. An extended light source shows the same effect as two point-like sources: you get overlapping fringes pattern behind the slits, the intensity pattern becomes blurry. So the light source has to be transformed to a point-like source. This happens by the help of an auxiliary slit between the light source and the slits.

enter image description here

For this picture Anna wrote: "Incandescent light is incoherent because it comes from many sources and the same is true for sunlight. By passing the light through the one slit he created a single coherent wave front." So my question stays unanswered. What makes the radiation behind a slit coherent?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    May 14, 2016 at 7:48

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A point source has a spherical wave front with the intensity falling as 1/r^2. The fronts are in constant phases because there is no dependence on theta and phi in the intensity.

For an aperture with a width see the question and answer here and links therein. It depends on the width of the slit, the frequency and the coherence length.

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  • $\begingroup$ Anna, I made an edit of my question. In addition there ist a fact, that for fringes there is no need in any slit or coherent source. Behind any obstacle a light source made fringes. For validation I refer to Michael Richmonds Diffraction effects during a lunar occultation: "When the Moon passes in front of a star, it also produces diffraction. But the situation is a bit different: this time, light passes through a "slit" which has only a single edge". $\endgroup$
    – HolgerFiedler
    Jan 16, 2016 at 18:36
  • $\begingroup$ @HolgerFiedler Light hitting an edge can mathematically approximated as as two point sources in the plane See everythingmaths.co.za/science/grade-11/06-2d-and-3d-wavefronts/… $\endgroup$
    – anna v
    Jan 16, 2016 at 18:48
  • $\begingroup$ Thanks for your calm. The link does not say anything about single edge and the mathematically approximated as two point sources. But please, let us stay with the main question about - rephrasing it - how a slit make a non coherent light source to wavefronts. $\endgroup$
    – HolgerFiedler
    Jan 16, 2016 at 19:00
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    $\begingroup$ ,In the plane perpendicular to the slit it is like two point sources, that is what the link draws. A single edge is like a point source . i.e. you could write the mathematics of a point source wave for the half hemisphere $\endgroup$
    – anna v
    Jan 16, 2016 at 19:13
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    $\begingroup$ It is the wave function of the photons. They are complex and when complex conjugated build up the electric magnetic and polarization of emergent light . see this maybe nist.gov/pml/div684/fcdc/upload/preprint.pdf arxiv.org/ftp/quant-ph/papers/0604/0604169.pdf and more on google search. The photon wave function is not much tooted because the classical EM is so successful in optics etc. $\endgroup$
    – anna v
    Jan 17, 2016 at 10:29
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Strictly, you can't just say the light behind a slit is coherent. You can say

  • it has a certain coherence time $\tau$, meaning it can interfere with a copy of itself which was delayed by time $\tau$

  • it has spatial coherence with respect to the light behind another slit, meaning they have a (somewhat) fixed phase difference and can interfere with each other

For temporal coherence, you illuminate the slit with a source like a continuous wave laser which has a narrow spectrum and long coherence time. Intuitively, it sends out looong wave packets so you can delay them a lot and they still overlap/interfere.

For spatial coherence, you take a small solid angle of some source and expand it, like you showed by passing a light source through a pinhole. Intuitively, you're making sure that you're grabbing one wavepacket at a time and making it interfere with itself, rather than grabbing two different packets whose phase relationship is random.

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Each of the two edges of a slit act as point sources and all light coming from them is coherent. This can easily be proven mathematically. Each edge forms a single edge fringe pattern on the screen with alternating dark and light fringes calculated as m=the square root of wavelength times distance times (m+3/4) where m = the number of spaces out from the edge. When the two single edge patterns overlap on the screen a single slit fringe pattern forms with equal spacing.See straight edge diffraction.

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Nothing about light traveling through a slit or opening actually makes spatially incoherent light become coherent. What occurs is that, where you have an incoherent light-source, i.e. a non-point or extended source, and you place in its path a small enough opening, you're isolating light that was emitted, relatively speaking, from a single point on that non-point source, and hence that is already relatively spatially coherent.

In the comments to the question, the OP has clarified that he's asking about some kind of interaction between the light and the electrons of the edge of the slit as the cause of the light becoming coherent after it leaves the slit. Such an idea, as can be seen, is very far from the truth.

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    $\begingroup$ How can it be far from the truth if all photons are emitted from electrons to begin with? $\endgroup$
    – Bill Alsept
    May 13, 2016 at 0:05
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    $\begingroup$ Even in a small enough opening if the light source is full spectrum you will still get different frequency photons or incoherent light. Just try shining white light through the experiment and you'll see the rainbow colors in the diffraction pattern $\endgroup$
    – Bill Alsept
    May 13, 2016 at 2:10
  • $\begingroup$ Actually, Bill, in your second comment you're exactly right. But we're talking about two different types of coherence. I'm talking about spatial coherence, whereas you're talking about temporal coherence. $\endgroup$
    – David Reishi
    May 13, 2016 at 2:16

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